Expert: Philip A. Stahl - 11/28/2006

Question
Hello,

Can you give assistance in solving the following? Thanks!

1) The lines of the Balmer series crowd close together as higher series members are considered. If each line were exactly one A (1 Angstrom) wide, how many Balmer lines would be individually visible without overlapping other lines?

2) The temperature of a gas is changed by a small amount. Derive an expression for the amount by which the electron pressure must be changed in order that the relative abundances of the first two ionization stages of a given element be unchanged.

3) By considering the first and third moments of tghe specific intensity:

I = I _o + I_1 cos (Q)

obtain the relation between the radiation pressure P(r) and the energy density, E.

Answer
Hello,

Here are suggested strategies, hints for solving the problems:

Problem 1

The Balmer lines are defined according to:

1/ L  = R (1/ 4 – 1/n^2)

where L defines the wavelength, n is an energy level > 2 and R is the Rydberg constant:

R = 1.097 x 10^7  m^-1


From this you should be able to obtain (with appropriate algebra, change of units):

L = 3645 [n^2/ (n^2 – 4)]

which yields all wavelengths in Angstroms, unlike the original expression which yields wavelength in meters. (Bear in mind here the need to pay attention to the units, since 1 A = 10^-10 m )


Overlapping of lines must begin for that ‘n’ for which we have the condition:

L(n) – L(n + 1) = 1 A

Now, consider L (wavelength) to be a function of n as given above –such that we have:

3645 L^-1 =  1 – 4n^-2

The next (and last) step in this solution hint, is that you have to make use of differentials (after differentiating) such that:

-3645 L^-2 dL  =  8n^-3  dn


From here, it is straightforward. Simply solve for n^3 on one side, with the derivative (-dn/dL) on the other.

You can then solve for n, and use the parameters of the problem which dictate: dn = 1 and dL = -1

This should work for you!



Problem 2

The free electrons – e.g. in an electron gas -  will produce a pressure given by:

P(e) = N(e) k T

Where k is the Boltzmann constant, k = 1.38 x 10^-23 J/K

Degrees of ionization are usually expressed by the partition function in logarithmic form:

log N(i+1)P(e)/ N(i) = -0.48 + log (2B(i+1)/B(i)) + 2.5 log T – 5040 I(i) / T

Where the second term on the right side bears the partition functions in terms of B

If these partition functions can be taken as constants then the above equation shows that ionization degrees will remain unchanged IF:

2.5 log T – 5040 I/ T – log P(e) = 2.5 log (T + dT) – 5040 I/ (T + dT) – log(P(e) + dP(e)

You should be able to show -  using appropriate substitutions - that the first equation can be reduced to the one above, using the assumption of the constant partition functions. (Note that the small d’s in the equation represent deltas  - not exact differentials!)

The preceding can be further reduced to:

log( 1 + dP(e)/ P(e)) =  2.5 log(1 + dT/ T) + 5040 I/ T  [dT/ T + dT]

Now, use the approximation such that for SMALL values of x, one has:

log(1+ x) = 0.4343 (x – x^2/ 2 + …..)

From this you should be able to derive the expression required, once you assume dP(e)/ P(e), and dT/T are very small.


Problem 3:

You have to start with the transfer equation for a spherically symmetric configuration, such that:

u(dI/dr) + (1 – u^2)/r (dI/dr) = -k (rho) I + j(rho)/ 4 pi

where u = cos Q  and du = -sin Q dQ

You should be able to demonstrate this from the geometry, e.g. using a long scalene triangle within which a perpendicular ‘drop down’ leaves angle Q between the two sides (r + dr) and r, and s forms the small leg segment at the very end.

In the smallest triangle formed, one has:  ds = dr/ cos Q and dr/ ds = cos Q

Further, dQ = - ds sin Q/ r

Now, for the first moment of the specific intensity I, you will hold r fixed and integrate the transfer eqn. with respect to du, over all space from 1 to -1.
In getting to the second moment eqn. you will see integrals of the form:

INT  (1 – u^2) dI

The hint here is to use integration by parts with dI = (1 – u^2) . Again, the limits of integration are from 1 to -1

This will lead to a divergence of flux equation

In getting to the third moment equation you will see integrals of form:

INT (u – u^3) dI

In the final hint, bear in mind the expansion of I in terms of moments is:

I = I_o + uI_1 + u^2I_2  + u^3 I_3 +  ……

And you will be interested in estimating the relative ratios of the coefficients for which,

[I_n/ I_n-1] << 1

If your integrations are correct you should find the energy density

E = 4 pi I_o/ c       

where c is the speed of light and I_o is the zeroth moment of I.

You should find for the radiation pressure:

P_r =  4 pi I_o/ 3c

And from these you ought to be able to obtain the specific relationship!